Consider an original ideal model with sample volume

V_{s}

, bath volume

V_{b}

, and initial sample solute concentration

c_{0}

. Assume each volume is fixed, a freely permeant, nonbinding solute, no chemical reactions, complete mixing on each side, and equal concentrations at equilibrium. Each fresh bath initially contains none of this solute. The protein is perfectly retained in this model.

The first equilibration redistributes the initial amount of substance

c_{0} V_{s}

through both compartments. Conservation gives

c_{0} V_{s} = c_{1} (V_{s} + V_{b}), c_{1} = c_{0} \frac{V _{s}}{V _{s} + V _{b}} .

This denominator includes the sample itself. For the chosen

V_{s} = 1 mL

and

V_{b} = 9 mL

,

c_{1} / c_{0} = 1/10

, rather than

1/9

. Nine-tenths of the initial solute is in the bath and one-tenth remains in the sample; nothing has vanished during equilibration.

Discarding the equilibrated bath removes its solute. A second fresh bath starts with only

c_{1} V_{s}

of solute in the entire system, so the same argument applies again. Induction gives the geometric progression

q = \frac{V _{s}}{V _{s} + V _{b}}, c_{n} = q c_{n - 1} = c_{0} q^{n}, \frac{c _{n}}{c _{0}} = 1 0^{- n} (V_{s} = 1, V_{b} = 9) .

Here

n

counts completed equilibrations with fresh baths:

n = 1

is the first bath,

n = 2

follows one replacement, and

n = 3

follows two replacements. The normalized sample concentrations are respectively

0.1

,

0.01

and

0.001

. Leaving the first bath in place for longer cannot lower the ideal equilibrium value below

0.1

.

The same conservation argument handles a fresh bath already containing a permeant solute at fixed concentration

b

:

c_{n} = \frac{V _{s} c _{n - 1} + V _{b} b}{V _{s} + V _{b}}, c_{n} - b = q (c_{n - 1} - b), c_{n} = b + (c_{0} - b) q^{n} .

Thus the sample approaches

b

, not necessarily zero. For example, if a hypothetical new buffer component starts at zero in the sample and at

b

in every bath, its sample concentration reaches

0.9 b

,

0.99 b

and

0.999 b

in the three equilibrations. Outward removal and inward replacement follow the same mathematics.

Equilibrium fractions do not specify elapsed time. An illustrative well-mixed linear ordinary differential equation model within one bath is

V_{s} \frac{d c _{s}}{d t} = - K (c_{s} - c_{b}), V_{b} \frac{d c _{b}}{d t} = K (c_{s} - c_{b}), K > 0.

Here

K

is an effective transfer coefficient with units of volume per time. Adding the two equations conserves

V_{s} c_{s} + V_{b} c_{b}

; subtracting their normalized forms gives

\frac{d ( c _{s} - c _{b} )}{d t} = - K (\frac{1}{V _{s}} + \frac{1}{V _{b}}) (c_{s} - c_{b}) .

The difference undergoes exponential decay. A bath replacement resets

c_{b}

, renewing the driving difference without making

c_{s}

jump instantly. The plotted points are equilibrium endpoints, not a measured time course or a timetable for an arbitrary protein.

Real sample volume can change through osmosis; proteins can be lost through leakage or adsorption to the semipermeable membrane. Consequently, a measured decrease in protein concentration alone cannot distinguish dilution from loss of protein. Measuring recovered volume as well as protein concentration permits an estimate of recovered protein mass.^[ref]

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